Problem 73
Question
Let \(P\) be a point on a ladder of length \(a+b, P\) being \(a\) units from the top end. As the ladder slides with its top end on the \(y\) -axis and its bottom end on the \(x\) -axis, \(P\) traces out a curve. Find the equation of this curve.
Step-by-Step Solution
Verified Answer
The path traced by point \(P\) is an ellipse given by \(\frac{x_1^2}{b^2} + \frac{y_1^2}{a^2} = 1\).
1Step 1: Understand the Problem
A ladder of length \(a+b\) rests on the coordinate plane, with its top on the \(y\)-axis and its bottom on the \(x\)-axis. We are tasked with finding the equation of the path traced by a point \(P\) which is \(a\) units from the top end as the ladder slides down.
2Step 2: Parametrize the Ladder
Let's consider the top of the ladder to be at \((0, y)\) and the bottom at \((x,0)\). The length of the ladder can be expressed using the Pythagorean theorem as \(x^2 + y^2 = (a+b)^2\). The point \(P\) can be parametrized as \((x_1, y_1)\) in terms of \((x,y)\).
3Step 3: Determine Coordinates of Point P
Since \(P\) is \(a\) units from the top, we consider it to be on the line joining the endpoints. Let's assume the coordinates of \(P\) are \(\left(\frac{bx}{a+b}, \frac{ay}{a+b}\right)\), a weighted average based on its division between the endpoints by \(a\) and \(b\).
4Step 4: Substitute into Circle Equation
Substitute \(x_1 = \frac{bx}{a+b}\) and \(y_1 = \frac{ay}{a+b}\) into the circle equation derived from the ladder's length. Therefore, \((x_1\cdot \frac{a+b}{b})^2 + (y_1\cdot \frac{a+b}{a})^2 = (a+b)^2\).
5Step 5: Simplify the Equation
From the substitution, the equation becomes \(\left( \frac{a+b}{b}x_1 \right)^2 + \left( \frac{a+b}{a}y_1 \right)^2 = (a+b)^2\), which simplifies to \(\frac{x_1^2}{b^2} + \frac{y_1^2}{a^2} = 1\).
6Step 6: Final Equation of the Path
The final equation \(\frac{x_1^2}{b^2} + \frac{y_1^2}{a^2} = 1\) describes an ellipse, which is the path traced by point \(P\) on the ladder as it slides.
Key Concepts
ParametrizationEllipse EquationCoordinate Geometry
Parametrization
In mathematics, parametrization is a technique used to express a set of coordinates or a geometric object in terms of parameters. It provides a way to describe complex geometrical configurations more simply by using one or more variables.When dealing with projectile motion or paths of objects, parametrization allows us to break movement into fundamental steps:
- Firstly, identify known points or boundaries. In our example, the ladder's length allows us to mark its endpoints on the coordinate plane.
- We then use parameters, say angles or lengths from end points, to express the location of a point.
Ellipse Equation
An ellipse is a geometric shape that appears as an elongated circle, defined mathematically by its unique properties and parameters. It represents a closed curve where the sum of distances from two distinct points, called foci, to any point on the ellipse is constant.In calculus and coordinate geometry, the standard equation for an ellipse centered at the origin is given as:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]Here, \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. When applied to the ladder problem, the equation:\[\frac{x_1^2}{b^2} + \frac{y_1^2}{a^2} = 1\]This reveals that the path traced by point \(P\) forms an ellipse relative to the coordinate axes. The essential idea is that as the ladder slides, the point follows a consistent elliptical trajectory described by the equation above.
Coordinate Geometry
Coordinate geometry, often known as analytic geometry, integrates algebra and geometry to determine the positions of points on a plane using an ordered pair of numbers. This helps in the mathematical plotting and manipulation of geometric forms based on numerical equations.The exercise with the ladder relies heavily on this concept. It involves:
- Assigning coordinates to the top and bottom of the ladder \((0, y)\) and \((x,0)\).
- Using fundamental relations like the Pythagorean theorem to express movements through equations \(x^2 + y^2 = (a+b)^2\).
Other exercises in this chapter
Problem 71
For an ellipse, let \(p\) and \(q\) be the distances from a focus to the two vertices. Show that \(b=\sqrt{p q}\), with \(2 b\) being the minor diameter.
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